Subject: Re: [xsl] Reverse order axis operators... From: David Carlisle <davidc@xxxxxxxxx> Date: Sat, 12 Jan 2002 16:35:09 GMT |
> All the reference material abd W3c reading says that these four Axis should > return node-list in reverse document order. They are reverse axese but that only means that they essentially collect their nodes in reverse order so ancestor::*[1] is the parent, ie the first ancestor going up. However the value of such an Xpath expression is a node set. Sets (unlike lists) do not have an order. So it does not matter whether you collect nodes using a forward or reverse axis or a combination of the two, the resuult of an expression is always an unordered set. When you come to iterate over that set using xsl:for-each the for-each sorts then into document order or some other order specified by xsl:sort but note it is the for-each (or copy-of) taht is ordering the set not that the set itself has any intrinsic order that "remembers" the order in which it was selected. David PS a)this is a FAQ I'm sure that the faq for this list has more info b) this is an XPath 1 answer. In the Xpath2 working draft (which is a very new draft and not in msxml) the semantix of these expressions are described in terms of sequences (ie ordered lists) rather than unordered sets, but the result is the same a for-each iterating over ancesto::* iterates over the nodes in document (forwards) order. _____________________________________________________________________ This message has been checked for all known viruses by Star Internet delivered through the MessageLabs Virus Scanning Service. For further information visit http://www.star.net.uk/stats.asp or alternatively call Star Internet for details on the Virus Scanning Service. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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